Cal's questions at Yahoo Answers regarding the derivative form of the FTOC

In summary, the derivative of a definite integral is equal to the negative of the integrand evaluated at the upper bound, multiplied by the derivative of the upper bound. This rule can also be applied to definite integrals with negative bounds by using the rule that the integral from a to b is equal to the negative of the integral from b to a.
  • #1
MarkFL
Gold Member
MHB
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Here are the questions:

How to get the derivative of a definite integral?

F(x) = the integral from x to 4 of sin(t^4) dt

F'(x) = ?

----

f(x) = the integral from x to 12 of t^7 dt

f'(x) = ?

I have several more like these. I've been working on them for 6 hours. None of my answers are right. I don't understand what I'm doing wrong!

I have posted a link there to this topic so the OP can see my work.
 
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  • #2
Hello Cal,

The derivative form of the Fundamental Theorem of Calculus states:

\(\displaystyle \frac{d}{dx}\int_a^{u(x)} f(t)\,dt=f\left(u(x) \right)\frac{du}{dx}\)

Another rule we will need is the following rule for definite integrals:

\(\displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\)

This rule can easily be demonstrated using the anti-derivative form of the Fundamental Theorem of Calculus:

\(\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)=-\left(F(a)-F(b) \right)=-\int_b^a f(x)\,dx\)

So, let's apply these rules to the given problems.

1.) \(\displaystyle F(x)=\int_x^4 \sin\left(t^4 \right)\,dt=-\int_4^x \sin\left(t^4 \right)\,dt\)

Hence:

\(\displaystyle F'(x)=-\frac{d}{dx}\int_4^x \sin\left(t^4 \right)\,dt=-\sin\left(x^4 \right)\)

2.) \(\displaystyle f(x)=\int_x^{12}t^7\,dt=-\int_{12}^x t^7\,dt\)

Hence:

\(\displaystyle f'(x)=-\frac{d}{dx}\int_{12}^x t^7\,dt=-x^7\)
 

FAQ: Cal's questions at Yahoo Answers regarding the derivative form of the FTOC

What is the derivative form of the Fundamental Theorem of Calculus (FTOC)?

The derivative form of the FTOC states that if a function f(x) is continuous on the interval [a, b] and differentiable on (a, b), then the definite integral from a to b of f'(x) is equal to f(b) - f(a).

How is the derivative form of the FTOC different from the original theorem?

The derivative form of the FTOC is a more specific version of the original theorem, as it only applies to continuous and differentiable functions. The original theorem applies to all continuous functions that have an antiderivative.

Can the derivative form of the FTOC be applied to both definite and indefinite integrals?

Yes, the derivative form of the FTOC can be applied to both definite and indefinite integrals. For definite integrals, it gives a specific value for the integral, while for indefinite integrals, it provides a general expression for the antiderivative.

How is the derivative form of the FTOC used in real-world applications?

The derivative form of the FTOC is commonly used in physics, engineering, and economics to solve problems involving rates of change and accumulation. It is also used in optimization problems to find the maximum or minimum value of a function.

Is it necessary to use the derivative form of the FTOC in all integration problems?

No, it is not necessary to use the derivative form of the FTOC in all integration problems. It is only applicable when the function is continuous and differentiable on the given interval. Otherwise, the original form of the FTOC or other integration techniques may be used.

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